Substituting (21) into (20) gives
dP x
=≡ = - +
dp x
Px
6p
x
dP
J dp
Substituting for dp
from (18) and using x = xs gives
dP
dP)
χ________1________Γs_ χ/ s _ (p-Px ʌ
xS- (x-x∖^-χs S rΛ6P m 6m6P)
X y x x χ x ^m x p p IIU
> 0.
(23)
Differentiating (6) w.r.t. x, we obtain
dP
dx)
dp
dx
x) dp 1 x
+ ~μ + — (P - P) + (P - P) —2 xP
x dx x (x)2
dp
dxP
-x
px
6P
x
dp
dxP
1 (P - P).
x
Substituting for dp∙ from (19) and using x = xs, we obtain
dP
dxP
S — (x — x)xm — xp
xp (6P
(p — p)p
m
6m6p)
< 0.
(24)
Proposition 3
Differentiating (8), and using Roy’s identity,
dv = Vpdp + vm dpp — Λ P = -vm dp (x — x) + x
dp> dp ∖Pp J dp>
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